Various Measures of Centrality By Marissa Stephens and Donna Choi - - PowerPoint PPT Presentation

Various Measures of Centrality

By Marissa Stephens and Donna Choi

Matrix of Family Relationships

F 1 F 2 F 3 F4 F5 F6 F7 F8 F9* F10 F11 F13 F14 F15 F16

1 1 1 1 2 1 2 1 2 1 2 1 1 1 2 1 1 1 1 1 2 2 1 1 1 2 1 1 1 2 1 1 1 2 2 1 1 1 2 2 1 1 1 1 1 2 1 1 1 1 1 1 2 1

* Family number 9 is the Medici Family

Map of Connections

1 16 15 14 12 13 2 11 3 10 4 9 8 7 6 5

Degree

 Degree of each family is a simple way of determining

centrality

 To find the number of degree per node (or family),

count how many relationships there are between that node and the other nodes.

 For Example, the Medici Family has a degree of 11

(The Medici Family is node number 9)

Degree of Each Family

=number of links per node

2 4 6 8 10 12 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Power

 A slightly more complicated way of calculating

centrality

 Takes into account people two relationships away  (like a friend of a friend)  Take the matrix M(people away a distance of 1) and

add it to M*M(people away an exact distance of 2).

 Sum each row in the resulting matrix to obtain the

power

Power of Each Family

=M+M^2

5 10 15 20 25 30 35 40 45 50 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Series2 Series1

In this case, Family number 3 has the most power

Markov Chains

 Illustrates transition diagrams of probability  Rows add up to 1  To transform an adjacency matrix into a Markov chain,

divide each element in a row by the row total

 This can be used to find the total fraction of influence

• f each family

 From now on, we will exclude family 12. It is isolated

from all of the other families.

Matrix of Markov Chain of Family Relationships

F 1 F 2 F 3 F4 F5 F6 F7 F8 F9 F10 F11 F13 F14 F15 F16

1 1/3 1/3 1/3 1/3 1/6 1/3 1/6 1/3 1/6 1/33 1/6 1/6 1/6 1/33 1/6 1/3 1/3 1/3 1/6 1/3 1/3 0.2 0.2 0.4 0.2 1/11 1/11 2/11 1/11 1/11 1/11 2/11 1/11 0.5 0.5 1/7 2/7 2/7 1/7 1/7 1/3 1/3 1/3 2/3 1/3 0.25 0.25 0.25 0.25 0.25 0.5 0.25

Fraction of Influence per Family

0.05 0.1 0.15 0.2 0.25 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Package Problem

Say one family wants to send a secret package to another family, but the first family can only send the package through people that family knows. If the package is randomly passed family to family, how long on average will it take the package to get to the desired family? Approach: Use a Markov chain matrix to determine the average number of transfers from one family.

Average number of passes

F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F13 F14 F15 F16

25 13 22 19 25 19 24 1 41 19 25 27 25 17 77 13 18 18 19 13 20 8 48 17 28 34 24 20 78 27 18 11 24 17 20 9 49 12 28 36 21 22 84 29 15 13 32 10 12 15 55 8 29 42 16 23 83 30 10 14 30 15 15 14 54 8 29 40 17 24 76 18 9 20 17 17 22 7 47 17 28 33 24 20 82 25 15 10 15 30 12 13 54 12 29 40 21 20 84 29 14 10 11 31 9 15 55 9 30 42 19 23 69 24 12 21 18 24 18 23 40 18 24 26 24 16 71 26 14 23 20 26 20 25 2 19 26 14 26 18 84 30 12 11 9 31 14 14 15 55 29 41 16 24 77 28 14 18 17 29 17 21 8 48 15 34 17 14 71 26 13 22 20 26 20 24 2 28 19 26 26 18 83 30 14 12 11 31 15 17 14 54 9 23 40 22 75 26 14 18 18 28 14 21 6 47 17 20 33 22

Average distance FROM each family

5 10 15 20 25 30

distance

Family 1 Family 2 Family 3 Family 4 Family 5 Family 6 Family 7 Family 8 Family 9 Family 10

Average distance TO each family

10 20 30 40 50 60 70 80 90 Average Family 1 Family 2 Family 3 Family 4 Family 5 Family 6 Family 7 Family 8 Family 9 Family 10 Family 11 Family 13 Family 14 Family 15 Family 16

Minimum Distance to Each Family

F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F13 F14 F15 F16

2 2 4 3 2 3 4 1 2 3 2 2 3 2 2 2 2 3 1 1 2 1 2 3 2 2 3 2 2 2 2 1 1 3 2 1 2 1 2 2 2 2 4 2 2 2 3 1 1 3 4 1 2 4 1 2 3 3 1 2 2 2 1 2 3 1 2 3 1 3 2 1 1 3 2 2 3 1 2 2 2 2 3 2 3 1 3 1 2 2 1 2 3 2 2 3 2 1 4 2 2 1 1 3 1 3 4 1 3 4 2 2 1 1 1 3 2 1 2 3 1 2 1 1 2 1 2 2 2 4 3 2 3 4 1 3 2 1 3 2 3 3 1 1 1 2 2 1 2 3 2 3 1 3 2 2 2 2 2 2 2 3 1 2 2 2 1 1 2 2 2 4 3 2 3 4 1 1 3 2 3 2 3 3 2 1 1 3 2 2 2 3 1 1 3 2 2 2 2 2 3 2 1 2 1 2 3 1 2 2

Family Distance from F9 F1

1

F2

1

F3

1

F4

3

F5

2

F6

1

F7

2

F8

3

F9 F10

1

F11

2

F13

1

F14

1

F15

2

F16

1

Family 9 would benefit most from forming connections with these families because they would minimize the distance between the two families. If a connection would be made with family 4 or 8, the shortest paths would be within a maximum of 2 links. The table on the right shows the minimum distance from family 9 after a connection to family 4 was added. Family Distance from F9 F1

1

F2

1

F3

1

F4

1

F5

2

F6

1

F7

2

F8

2

F9 F10

1

F11

2

F13

1

F14

1

F15

2

F16

1

Most Used Paths

 Big Question:

What connections are used most frequently to obtain the shortest path?

 Approach:

Using the shortest paths, count the number of times a connection ismade.

Path Usage for Shortest Path (from row to column)

F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F13 F14 F15 F16

14 14 13 12 13 13 12 12 11 11 15 11 13 14 13 15 12 13 17 15 11 13 12 12 16 12 14 15 14 16 13 15 18 15 12 14 13 14 17 13 13 16 12 12 12 8 13 12 11 9 11 10 13 11 12 12 12 14 15 9 11 9 11 7 11 10 13 11 12 14 16 18 21 18 14 15 14 16 15 13 19 15 15 18 13 16 15 13 9 12 12 8 12 8 14 12 10 16 11 11 11 11 10 12 11 8 10 9 14 10 8 11 13 15 15 9 8 14 9 10 12 7 16 12 9 15 11 13 13 12 11 12 12 11 11 10 14 13 12 13 11 13 14 11 10 10 8 10 6 10 12 10 11 13 15 17 17 14 13 16 14 15 15 14 12 14 17 20 11 13 13 12 11 12 12 11 11 13 10 14 12 13 12 14 12 12 11 13 9 8 7 11 10 16 11 12 14 16 16 13 14 15 16 12 14 13 13 20 13 13

The most used path is from Family 6 to Family 3

DeGroot Model

 This demonstrates what percent of a decision will

belong to each family.

 Uses Markov chains to determine how a consensus will

be reached.

 Solve ∏ *T= ∏

 ∏= the static constant of the Markov chain

DeGroot percentage (own opinion excluded)

1% 3% 7% 10% 8% 3% 9% 8% 15% 3% 10% 0% 5% 4% 6% 7% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

DeGroot Percentage

(considering own opinion as ½ of influence)

1% 2% 6% 10% 7% 4% 9% 6% 27% 3% 9% 0% 4% 5% 6% 4% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Life without the Medici

 Deleting the Medici Family form the Markov Chain yields:

F1 F2 F3 F4 F5 F6 F7 F8 F10 F11 F13 F14 F15 F16

0.5 0.5 0.5 0.25 0.25 0 0.333333 0.166667 0 0.333333 0 0.166667 0 0.333333 0 0.166667 0 0.333333 0 0.166667 0.5 0.5 0 0.166667 0 0.333333 0 0.333333 0 0.166667 0.2 0.2 0.4 0.2 1 0 0.142857 0.285714 0.285714 0 0.142857 0 0.142857 0.5 0.5 1 0.25 0.25 0.25 0.25 0.5 0.5

Graph of Relationships Without the Medici

1 16 15 14 12 13 2 11 3 10 8 7 6 4 5

Distance FROM each family (no Medici)

2 4 6 8 10 12 14 16 18 20

average distance

average distance

Distance TO each family (no Medici)

5 10 15 20 25 30 average distance Family 2 Family 3 Family 4 Family 5 Family 6 Family 7 Family 8 Family 10 Family 11 Family 13 Family 14

Without the Medici

 Family 1 becomes completely disconnected  The group is no longer connected  Families 10 and 14 are isolated from the rest of the

families

 Packages cannot be sent between certain families  Eleven links are destroyed  Therefore, the Medici family is a critical point in the

Renaissance family social group.

…And this all came from a single 16X16 Matrix